PRACTICE QUESTIONS ON COMBINATIONS

Question 1 :

If ⁽ⁿ⁺¹⁾C₈ : ⁽ⁿ⁻³⁾ P₄ = 57 : 16, find the value of n. 

Solution :

⁽ⁿ⁺¹⁾C₈  =  (n + 1)!/(n + 1 - 8)! 8! ==>  (n + 1)!/(n - 7)! 8!  --(1)

⁽ⁿ⁻³⁾ P₄ =  (n - 3)!/(n - 3 - 4)!  ==>  (n - 3)!/(n - 7)! --(2)

(1) : (2)

[(n + 1)!/(n - 7)! 8!] : [(n - 3)!/(n - 7)! ]  =  57 : 16

(n + 1)n(n - 1)(n - 2)/(8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2)  =  57 / 16

(n + 1) n (n - 1)(n - 2)  =  (57/16) ⋅ (8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2)

(n + 1) n (n - 1)(n - 2)  =  (57 ⋅  7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3) 

(n + 1) n (n - 1)(n - 2)  =  (3 ⋅ 19 ⋅  7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3) 

(n + 1) n (n - 1)(n - 2)  =  21 ⋅ 20 ⋅  19 ⋅ 18

n  =  20

Hence the value of n is 20.

Question 2 :

Prove that ²ⁿC_n = [2ⁿ × 1 × 3 × ·· · (2n − 1)] / n!.  

Solution :

L.H.S

  =  ²ⁿC_n  =  2n!/(2n-n)! n!  =  2n!/n! n!

  =  [2n (2n - 1)(2n - 2) .............  ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1] / n! n!

  =  [ 1 ⋅ 3 ⋅ 5 ⋅.............(2n - 1) ] [2 ⋅ 4 ⋅ 6 ⋅.......2n] / n! n!

  =  [ 1 ⋅ 3 ⋅ 5 ⋅.............(2n - 1) ] 2ⁿ [1 ⋅ 2 ⋅ 3 ⋅.......n] / n! n!

  =  [ 1 ⋅ 3 ⋅ 5 ⋅.............(2n - 1) ] 2ⁿ n! / n! n!

  =   2ⁿ [ 1 ⋅ 3 ⋅ 5 ⋅.............(2n - 1)] / n! ---> R.H.S

Hence it is proved.

Question 3 :

Prove that if 1 ≤ r ≤ n then n × ⁽ⁿ⁻¹⁾ C_r−1 = (n − r + 1) ⁿC_r−1.

Solution :

Question 4 :

Kabaddi coach has 14 players ready to play. How many different teams of 7 players could the coach put on the court?

Solution :

Number of ways of selecting 7 players out of 14 players 

=  ¹⁴C₇ 

=  14! / (14 - 7)! 7! 

=  14! / 7! 7!

=  (14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7!)/7! 7!

=  (14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8) / 7!

=  (14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8) / (7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2)

=  3432

Question 5 :

There are 10 points in a plane, out of which 4 points are collinear. The number of triangles formed with vertices at these points is

(a) 20        (b) 120       (c) 116      (d) none of these

Solution :

Number of points in the plane = 10

To create a triangle we need to connect 3 points.

= ¹⁰C₃ - ⁴C₃

= 10!/(10 - 3)! 3 ! - 4

 = 10!/7! 3! - 4

= (10 x 9 x 8 x 7!)/(7! x 3 x 2 x 1) - 4

= 10 x 3 x 4 - 4

= 120 - 4

= 116

So, the required number of ways is 116.

Question 6 :

10 students are participating in a competition. In how many different ways can the first prize be won? (There are 3 prizes)

(a) 720     (b) 60        (c) 30       (d) 120

Solution :

This is not selection, so we have to use the concept of permutation.

Total number of ways to win the first prize = ¹⁰ P ₃

= 10!/(10 - 3)!

= 10 x 9 x 8 x 7 !/7!

= 10 x 9 x 8

= 720

Question 7 :

Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to

(a) 60     (b) 120      (c) 7200       (d) 720

Solution :

Total number of vowels = 4

Total number of consonants = 5

Number of vowels to be chosen = 2

Number of consonants to be chosen = 3

Total number of words = ⁴C₂ x ⁵C₃ x (2 + 3)!

= (4!/2! 2!) x (5!/2! 3!) x 5! 

= 6 x 10 x 5!

= 6 x 10 x 120

= 7200

So, the answer is option c.

Question 8 :

All the letters of the word ‘EAMCOT’ are arranged in different possible ways. The number of such arrangements in which no two vowels are adjacent to each other is

(a) 360      (b) 144      (c) 72     (d) 54

Solution :

In the word EAMCOT,

Consonants = 3

vowels = 3

Since no two vowels have to be together. Possible ways may be VMVCVTV

⁴P₃, 3 consonants can be arranged in 3! = 6 ways

So, total number of ways

= 4P3 × 6 = 4 × 3 × 2 × 6 = 144

Question 9 :

The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is

(a) 105     (b) 15    (c) 175     (d) 185

Solution :

Required number of triangle 

= ¹²C₃ - ⁷C₃

= 12!/9!3! - 7!/4! 3!

= (12 x 11 x 10 x 9!)/9! x 3! - (7 x 6 x 5 x 4!)/4! 3!

= 2 x 11 x 10 - 7 x 5

= 220 - 35

= 185

So, option d is correct.

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